You are Here: Home >< Maths

# Complex number transformations involving arguments

Announcements Posted on
TSR's new app is coming! Sign up here to try it first >> 17-10-2016
1. https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

Q12) How can z be transformed onto a full circle? The circle equation I get but doesn't it have to satisfy 0 < x < 1 and -1 < y < 0?
Zacken TeeEm
Can you be more specific? I don't fully understand your question. Where does 0 < x < 1 and -1 < y < 0 come from?
3. (Original post by Zacken)
Can you be more specific? I don't fully understand your question. Where does 0 < x < 1 and -1 < y < 0 come from?
So instead of treating arg(z) = pi/4 as x + xi the MS suggested, I did it the argument way and got arg (1 - wi) - arg (w -1) = pi/4, which is a circle that satisfies mod w = 1, however only within the ranges I posted above. Surely the MS method is wrong as it treats arg (z) = pi/4 as x + xi, which is only true for values in the positive quadrant.. hope that makes more sense..?
So instead of treating arg(z) = pi/4 as x + xi the MS suggested, I did it the argument way and got arg (1 - wi) - arg (w -1) = pi/4, which is a circle that satisfies mod w = 1, however only within the ranges I posted above. Surely the MS method is wrong as it treats arg (z) = pi/4 as x + xi, which is only true for values in the positive quadrant.. hope that makes more sense..?
But arg(z) = pi/4 is only values in the positive x and y quadrant. That's why it's called a half-line.

Anyways, with your method, why does arg(1-wi) - arg(w-1) bring up the ranges that you specified?
5. (Original post by Zacken)
But arg(z) = pi/4 is only values in the positive x and y quadrant. That's why it's called a half-line.

Anyways, with your method, why does arg(1-wi) - arg(w-1) bring up the ranges that you specified?

Does that help?

Does that help?
Question doesn't say the image of the given half line is the circle, only that the image of the half line lies on the circle - this doesn't imply the entire circle.
7. (Original post by ghostwalker)
Question doesn't say the image of the given half line is the circle, only that the image of the half line lies on the circle - this doesn't imply the entire circle.
I realise that, but am I right in saying that the complex numbers on arg(z) = pi/4 can only be mapped onto those in the shaded region I previously specified?
I realise that, but am I right in saying that the complex numbers on arg(z) = pi/4 can only be mapped onto those in the shaded region I previously specified?
Yes, it's just the lower right quadrant, but it's the part of the circle only, not any area.
9. (Original post by ghostwalker)
Yes, it's just the lower right quadrant, but it's the part of the circle only, not any area.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: June 3, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams